Bayes theorem involving drug testing A drugs test for athletes is 99% reliable:  applied to a drug taker, it gives a positive result 99% of the time;given to a non-taker, it gives a negative result 99% of the time.  It is estimated that 1% of athletes take drugs.What fraction of randomly tested athletes fail the test?
I am completely lost on how to do this as i wasn't in class that day so how would I use bayes theorem to solve this? 
Thanks
 A: Lets give the test to $10,000$ people.  
$9,900$ people are not drug users (drug free).  $1\%$ of them score false positives.
$100$ are drug users, $99\%$ are detected.  $1\%$ are false negatives.
\begin{matrix}&\text{Pass}&\text{Fail}\\
\text{Drug Free}& 9,801& 99\\
\text{Drug User} &1& 99\end{matrix}
$198$ fail the test.  $\frac 12$ of those that fail are false positives and $\frac 12$ are true positives.
A: Testing a random athlete: let A denote a positive test result, B the guy doing drugs. Then:$\mathsf P(A\mid B)=0.99, \mathsf P(A^{\complement}\mid B^{\complement})=0.99$ with $A^{\complement}$ being the complement of $A$, thus a negative result with $\mathsf P(A^{\complement})=1-\mathsf P(A)$ and analogous $B^{\complement}$. 
You are also given $\mathsf P(B)=0.01$ 
You are searching for $\mathsf P(A)=\mathsf P(A\cap B)+\mathsf P(A\cap B^{\complement})$ the first is obtained trivially from the defintion of conditional probability, the second equaling $\mathsf P(A\mid B^{\complement})\cdot\mathsf P(B^{\complement})$ which are both avaible as $1-\mathsf P(A^{\complement}\mid B^{\complement})$ and $1-\mathsf P(B)$
